Problem: Assume that $S$ is an outwardly oriented, piecewise-smooth surface with a piecewise-smooth, simple, closed boundary curve $C$ oriented positively with respect to the orientation of $S$. $ \oint_C (4y \hat{\imath} + z\cos(x) \hat{\jmath} - y \hat{k}) \cdot dr$ Use Stokes' theorem to rewrite the line integral as a surface integral. $ \iint_S ( $ $ \hat{\imath} + $ $\hat{\jmath} + $ $ \hat{k} ) \cdot dS$
Explanation: Assume we have a continuously differentiable three-dimensional vector field $F(x, y, z)$, an oriented piecewise-smooth surface $S$, and a piecewise-smooth, simple, closed boundary curve $C$ oriented positively with respect to $S$. Then Stokes' theorem states that we have the equality below: $ \oint_C F \cdot dr = \iint_S \text{curl}(F) \cdot dS$ If $C$ is negatively oriented, the line integral is equal to the negative of the double integral. [What does any of that mean?] When we use Stokes' theorem to translate from line integrals to surface integrals, we know $F$ and we want to find $\text{curl}(F)$. $\begin{aligned} F(x, y, z) &= 4y \hat{\imath} + z\cos(x) \hat{\jmath} - y \hat{k} \\ \\ \text{curl}(F) &= \det \begin{pmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ \\ \dfrac{\partial}{\partial x} & \dfrac{\partial}{\partial y} & \dfrac{\partial}{\partial z} \\ \\ 4y & z\cos(x) & -y \end{pmatrix} \\ \\ &= (-1 - \cos(x)) \hat{\imath} \\ \\ &+ (-z\sin(x) - 4)\hat{k} \end{aligned}$ Now that we know $\text{curl}(F)$, we can use it to find a surface integral equivalent to the original line integral: $ \iint_S \left[ (-1 - \cos(x))\hat{\imath} + (0)\hat{\jmath} + (-z\sin(x) - 4) \hat{k} \right] \cdot dS$